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S_A_V [24]
3 years ago
10

There are 12 boxes at a dock. 1 is yours with stuff in it and you have 11 others that have products in them, which have all the

same weight, but you dont know if the 11 boxes separately have more weight or less weight than yours. You have a scale that has 3 uses to weigh them. A robot weighs the boxes so you cant feel the weight. How do you figure out which one is yours?
Mathematics
1 answer:
sammy [17]3 years ago
5 0

Answer:

take a random box and just hope its yours

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Fabian harvests 10 pounds of tomatoes from his garden. He needs 225 pounds to make a batch of soup. If he sets aside 2.8 pounds
nadya68 [22]
He can make 3 batches of soup.

Starting with 10 pounds of tomatoes and taking out the number of pounds he sets aside for spaghetti sauce, 2.8, we have:
10-2.8 = 7.2 pounds left.

0.2 is read as "two tenths," so 7.2 = 7 2/10

We divide this remaining number of pounds by 2 2/5 for each batch of soup:

7 2/10 ÷ 2 2/5

Convert each to an improper fraction:
72/10 ÷ 12/5

Multiply by the reciprocal:
72/10 × 5/12 = 360/120 = 3


5 0
3 years ago
Double a number is equal to the number decreased by five?
kotegsom [21]
2n=n-5  subtract n from both sides

n=-5

check...

2(-5)=(-5)-5

-10=-10
6 0
3 years ago
3 Which of the following represents a function? A {(-2, 1), (-2, 0), (-2, 1), (-2, 2)} B {(-1,0), (0, 1), (1, 2), (2, 3)} C {(1,
solong [7]

Answer:

A

Step-by-step explanation:

because they want you to look for the number who repeat in the right like A (

4 0
3 years ago
Read 2 more answers
Suppose f (x) = x^2. Find the graph of f(x+2)
NISA [10]
\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)={{  A}}({{  B}}x+{{  C}})+{{  D}}
\\\\
~~~~y={{  A}}({{  B}}x+{{  C}})+{{  D}}
\\\\
f(x)={{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\\\
f(x)={{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\\\
f(x)={{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\\\\
--------------------

\bf \bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
~~~~~~\textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
~~~~~~if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
~~~~~~if\ {{  D}}\textit{ is negative, downwards}\\\\
~~~~~~if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}

with that template in mind,

\bf f(x)=x^2\qquad \quad f(x+2)=(x+2)^2\implies f(x+2)=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+2})^2\stackrel{D}{+0}

C = 2         B = 1        C/B = 2/1 or +2,    horizontal left shift of 2 units

f(x) shifted left by 2 units is f(x+2).
8 0
3 years ago
2 ( −14 + r ) − ( −3r − 5 )
svet-max [94.6K]
-28+2r+3r+5 
simplified it is -23+5r
4 0
3 years ago
Read 2 more answers
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